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A007863
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Number of hybrid binary trees with n internal nodes.
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28
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1, 2, 7, 31, 154, 820, 4575, 26398, 156233, 943174, 5785416, 35955297, 225914342, 1432705496, 9158708775, 58954911423, 381806076426, 2485972170888, 16263884777805, 106858957537838, 704810376478024, 4664987368511948, 30974829705533240, 206266525653071416
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OFFSET
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0,2
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COMMENTS
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Definition of hybrid binary trees:
An (a,n)-labeled binary tree is a binary tree where each internal node is labeled by "a" (for associative) or "n" (for nonassociative). We define on the set of (a,n)-labeled binary trees with a given number of nodes an equivalence relation as follows: denote a tree with a root labeled "a" with left subtree A and right subtree B by AaB. Then we declare the trees (AaB)aC and Aa(BaC) equivalent, and two trees are equivalent if and only if one can go from one to the other by doing such transformations within any of their subtrees.
A hybrid binary tree is an equivalence class of (a,n)-labeled binary trees under this relation. (End)
Also the number of Dyck n-paths with up steps colored in two ways (N or A) and avoiding AA. The 7 Dyck 2-paths are NDND, NDAD, ADND, ADAD, NNDD, NADD, and ANDD. - David Scambler, May 21 2012
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LINKS
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FORMULA
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G.f. A(x) satisfies: x^2*A(x)^3 + x*A(x)^2 + (-1+x)*A(x) + 1 = 0.
a(n) = 3F2({-n, n+1, n+2 } ; {1, 3/2})( -(1/4) ). - Olivier Gérard, Apr 23 2009
a(n) = (1/(n+1))*Sum_{k=0..n} binomial(n+k,n)*binomial(n+k+1,n-k). - Vladimir Kruchinin, Dec 24 2010
G.f.: A(x) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*A(x)^k] * x^n/n ). - Paul D. Hanna, Feb 13 2011
The formal inverse of the g.f. A(x) is (sqrt(5*x^2 - 2*x + 1) - (1+x))/(2*x^2). - Paul D. Hanna, Aug 21 2012
The radius of convergence of g.f. A(x) is r = 0.1407810125... with A(r) = 2.1107712092... such that y=A(r) satisfies 5*y^3 - 12*y^2 + 4*y - 2 = 0. - Paul D. Hanna, Aug 21 2012
D-finite with recurrence: 45*n*(n+1)*a(n) - 2*n*(157*n-71)*a(n-1) + 12*(-3*n^2+15*n-14)*a(n-2) + 2*(-14*n^2+43*n-21)*a(n-3) - 4*(n-3)*(2*n-7)*a(n-4) = 0. - R. J. Mathar, Jun 03 2014
Recurrence (of order 3): 5*n*(n+1)*(35*n-62)*a(n) = 6*n*(210*n^2 - 477*n + 181)*a(n-1) - 4*n*(35*n^2 - 132*n + 115)*a(n-2) + 2*(n-2)*(2*n-5)*(35*n-27)*a(n-3). - Vaclav Kotesovec, Jul 11 2014
a(n) ~ sqrt((s*(1+s+2*r*s^2))/(1+3*r*s)) / (2*sqrt(Pi) * r^n * n^(3/2)), where r = 52/(3*(181 + 105*sqrt(105))^(1/3)) - 1/6*(181 + 105*sqrt(105))^(1/3) + 1/3 = 0.1407810125885522212..., s = 1/15*(12 + (1323 - 105*sqrt(105))^(1/3) + (21*(63 + 5*sqrt(105)))^(1/3)) = 2.110771209224758867... . - Vaclav Kotesovec, Jul 11 2014
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EXAMPLE
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G.f. = 1 + 2*x + 7*x^2 + 31*x^3 + 154*x^4 + 820*x^5 + 4575*x^6 + ...
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MAPLE
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A:= proc(n) option remember; if n=0 then 1 else convert(series((x^2 *A(n-1)^3 +x*A(n-1)^2 +1)/ (1-x), x=0, n+1), polynom) fi end: a:= n-> coeff(A(n), x, n): seq(a(n), n=0..30); # Alois P. Heinz, Aug 22 2008
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MATHEMATICA
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InverseSeries[Series[(y-y^2-y^3)/(1+y), {y, 0, 24}], x] (* then A(x)=y(x)/x . - Len Smiley, Apr 14 2000 *)
Table[ HypergeometricPFQ[{-n, 1 + n, 2 + n}, {1, 3/2}, -(1/4)], {n, 0, 20}] - Olivier Gérard, Apr 23 2009
a[ n_] := If[ n < 0, 0, HypergeometricPFQ[{-n, 1 + n, 2 + n}, {1, 3/2}, -1/4]]; (* Michael Somos, Dec 31 2014 *)
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PROG
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(Macsyma) taylor_solve_choose_order:true; taylor_solve( A^3*X^2+A^2*X+A*(X-1)+1, A, X, 0, [ 20 ]);
(PARI) {a(n) = if( n<0, 0, sum(k=0, n, binomial(n+k, n) * binomial(n+k+1, n-k)) / (n+1))};
(PARI) {a(n) = local(A = 1 + x + x * O(x^n)); for(i=1, n, A = 1 + x * (A + A^2) + x^2 * A^3); polcoeff(A, n)};
(PARI) {a(n) = local(A=1+x); for(i=1, n, A = exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2 * (A + x * O(x^n))^j) * x^m / m))); polcoeff(A, n, x)};
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Jean Pallo (pallo(AT)u-bourgogne.fr)
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STATUS
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approved
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